Question: Let $a,b,c$ be positive real numbers such that $a+b+c=10$ and $ab+bc+ca=25$.  Let $m=\min\{ab,bc,ca\}$.  Find the largest possible value of $m$.
Explanation: The given conditions are symmetric in $a,$ $b,$ and $c,$ so without loss of generality, we can assume that $a \le b \le c.$  Then $10 = a + b + c \le 3c,$ so $c \ge \frac{10}{3}.$  By AM-GM,
\[(a + b)^2 \ge 4ab.\]Then
\[(10 - c)^2 \ge 4(25 - ac - bc) = 100 - 4(a + b)c = 100 - 4(10 - c)c.\]This reduces to $3c^2 - 20c = c(3c - 20) \ge 0,$ so $c \le \frac{20}{3}.$

Now,
\[m = \min\{ab,ac,bc\} = ab = 25 - c(a + b) = 25 - c(10 - c) = (c - 5)^2.\]Since $\frac{10}{3} \le c \le \frac{20}{3},$ $m = ab \le \frac{25}{9}.$

Equality occurs when $a = b = \frac{5}{3}$ and $c = \frac{20}{3},$ so the maximum value of $m$ is $\boxed{\frac{25}{9}}.$